Is it always possible to partition $\mathbb{R}$ (or any other standard Borel space) into precisely $\aleph_1$ Borel sets?

On the one hand, this is trivial if the Continuum Hypothesis holds. Less trivially, this also follows from $\mathrm{cov}(\mathcal{M}) = \aleph_1$, $\mathrm{cov}(\mathcal{N}) = \aleph_1$, $\mathfrak{d} = \aleph_1$, and similar hypotheses. However, I can't think of a general argument that allows one to split $\mathbb{R}$ into precisely $\aleph_1$ pairwise disjoint nonempty Borel pieces.

On the other hand, PFA or MM might give a negative answer but I don't see a good handle from that end either.

I don't have the time right now to check the details, but it looks like you can build such a partition out of a Hausdorff gap. Of course Silver's theorem rules out the partition coming from a Borel equivalence relation (if CH fails).
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Clinton ConleyAug 6 '11 at 20:12

Thanks, Clinton. I thought of using a Hausdorff gap on the walk home. I think it works too.
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François G. Dorais♦Aug 6 '11 at 20:29

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I think it is not well understood which cardinals $\kappa$ have the property that the real line can be partitioned into $\kappa$ Borel sets. As Andreas points out, this is known in the case $\kappa = \aleph_1$. I think it is open, however, whether the continuum being at least $\aleph_n$ implies that the real line can be partitioned into $\aleph_n$ Borel sets (for $n > 1$). If I remember correctly, some form of this problem is due to Steve Watson.
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Justin MooreAug 6 '11 at 22:37

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Just to point out, the question I asked as "A compactness property for Borel sets" (not sure how to make a link - sorry) is (via easy reasoning) the negation of the question asked here. So all the constructions given as negative answers to my question can be imported as positive answers to this one. Similarly, Andreas Blass' very simple (pace the classical theorem) answer to this question, provides yet another counterexample to my question.
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Alex SimpsonAug 8 '11 at 9:38

3 Answers
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It suffices to express $\mathbb R$ as the union of $\aleph_1$ (not necessarily disjoint) Borel sets such that no countably many of them cover $\mathbb R$, because then you can list them in an $\omega_1$-sequence and subtract from each one the union of the previous ones. Partition $\mathbb R$ into a non-Borel $\Pi^1_1$ set $A$ (say the set of codes of well-orderings of $\omega$) and its complement. A classical theorem says that any $\Pi^1_1$ set is a union of $\aleph_1$ Borel sets, and so is every $\Sigma^1_1$ set. Apply that to $A$ and to $\mathbb R-A$ to get $\mathbb R$ as a union of $\aleph_1$ Borel sets. No countably many of them cover $\mathbb R$ because $A$ is not Borel and thus not a countable union of Borel sets.

Thanks Andreas! This is much more constructive than I thought it would be.
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François G. Dorais♦Aug 6 '11 at 20:31

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A slightly more explicit version of Andreas' argument: First, identify reals with trees in some Borel way. Here, trees are subsets of $\omega^{< \omega}$ which are downward closed. Next, each tree $T$ has a derivative $T'$: remove all leaves. This naturally defines a transfinite sequence of iterated derivatives $(T^{(\alpha)}: \alpha< \omega_1)$. Let the ("Cantor-Bendixson") rank of $T$ be the first $\alpha<\omega_1$ such that $T^{(\alpha)}$ has no leaves (possibly, but not necessarily, because it is empty). Now the set of trees with a given rank is certainly Borel.
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GoldsternFeb 21 '12 at 20:28

There is a Hausdorff gap, a sequence $\{f_\alpha,g_\alpha:\alpha\lt\omega_1\}$ of $\omega\to\omega$ functions such that $f_\alpha\lt^*f_\beta<^*g_\beta<^*g_\alpha$ hold for $\alpha\lt\beta\lt\omega_1$ (here $f\lt^* g$ denotes eventual dominance, i.e., that $f(n)\lt g(n)$ holds for large $n\lt\omega$) and there is no function $f:\omega\to\omega$ such that
$f_\alpha\lt^* f\lt^* g_\alpha$ holds for $\alpha\lt\omega_1$. If we identify the reals with ${}^\omega\omega$ and set $H_\alpha=\{f:f_\alpha\lt^* f \lt^* g_\alpha\}$ then $\{H_\alpha:\alpha\lt\omega_1\}$ is a decreasing sequence of nonempty $F_\sigma$ with empty intersection.